\(\int \frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {e x} (c+d x)^{3/2}} \, dx\) [898]

Optimal result

Integrand size = 36, antiderivative size = 149 \[ \int \frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {e} \sqrt {b c^2-b d^2 x^2}}{\sqrt {b} \sqrt {d} \sqrt {e x} \sqrt {c+d x}}\right )}{\sqrt {d} \sqrt {e}}-\frac {2 \sqrt {2} \sqrt {b} \arctan \left (\frac {\sqrt {e} \sqrt {b c^2-b d^2 x^2}}{\sqrt {2} \sqrt {b} \sqrt {d} \sqrt {e x} \sqrt {c+d x}}\right )}{\sqrt {d} \sqrt {e}} \] Output:

2*b^(1/2)*arctan(e^(1/2)*(-b*d^2*x^2+b*c^2)^(1/2)/b^(1/2)/d^(1/2)/(e*x)^(1 
/2)/(d*x+c)^(1/2))/d^(1/2)/e^(1/2)-2*2^(1/2)*b^(1/2)*arctan(1/2*e^(1/2)*(- 
b*d^2*x^2+b*c^2)^(1/2)*2^(1/2)/b^(1/2)/d^(1/2)/(e*x)^(1/2)/(d*x+c)^(1/2))/ 
d^(1/2)/e^(1/2)
 

Mathematica [A] (verified)

Time = 4.48 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\frac {2 \sqrt {x} \sqrt {b \left (c^2-d^2 x^2\right )} \left (\sqrt {c} \sqrt {1-\frac {d x}{c}} \arcsin \left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}}\right )-\sqrt {2} \sqrt {c-d x} \arctan \left (\frac {\sqrt {2} \sqrt {d} \sqrt {x}}{\sqrt {c-d x}}\right )\right )}{\sqrt {d} \sqrt {e x} (-c+d x) \sqrt {c+d x}} \] Input:

Integrate[Sqrt[b*c^2 - b*d^2*x^2]/(Sqrt[e*x]*(c + d*x)^(3/2)),x]
 

Output:

(2*Sqrt[x]*Sqrt[b*(c^2 - d^2*x^2)]*(Sqrt[c]*Sqrt[1 - (d*x)/c]*ArcSin[(Sqrt 
[d]*Sqrt[x])/Sqrt[c]] - Sqrt[2]*Sqrt[c - d*x]*ArcTan[(Sqrt[2]*Sqrt[d]*Sqrt 
[x])/Sqrt[c - d*x]]))/(Sqrt[d]*Sqrt[e*x]*(-c + d*x)*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {586, 140, 27, 65, 104, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {e x} (c+d x)^{3/2}} \, dx\)

\(\Big \downarrow \) 586

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \int \frac {\sqrt {b c-b d x}}{\sqrt {e x} (c+d x)}dx}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 140

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\int \frac {2 b c}{\sqrt {e x} (c+d x) \sqrt {b c-b d x}}dx-b \int \frac {1}{\sqrt {e x} \sqrt {b c-b d x}}dx\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (2 b c \int \frac {1}{\sqrt {e x} (c+d x) \sqrt {b c-b d x}}dx-b \int \frac {1}{\sqrt {e x} \sqrt {b c-b d x}}dx\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (2 b c \int \frac {1}{\sqrt {e x} (c+d x) \sqrt {b c-b d x}}dx-2 b \int \frac {1}{\frac {b d x e}{b c-b d x}+e}d\frac {\sqrt {e x}}{\sqrt {b c-b d x}}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (4 b c \int \frac {1}{c e+\frac {2 b c d x e}{b c-b d x}}d\frac {\sqrt {e x}}{\sqrt {b c-b d x}}-2 b \int \frac {1}{\frac {b d x e}{b c-b d x}+e}d\frac {\sqrt {e x}}{\sqrt {b c-b d x}}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {2 \sqrt {2} \sqrt {b} \arctan \left (\frac {\sqrt {2} \sqrt {b} \sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {b c-b d x}}\right )}{\sqrt {d} \sqrt {e}}-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {b c-b d x}}\right )}{\sqrt {d} \sqrt {e}}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

Input:

Int[Sqrt[b*c^2 - b*d^2*x^2]/(Sqrt[e*x]*(c + d*x)^(3/2)),x]
 

Output:

(Sqrt[b*c^2 - b*d^2*x^2]*((-2*Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt[d]*Sqrt[e*x])/( 
Sqrt[e]*Sqrt[b*c - b*d*x])])/(Sqrt[d]*Sqrt[e]) + (2*Sqrt[2]*Sqrt[b]*ArcTan 
[(Sqrt[2]*Sqrt[b]*Sqrt[d]*Sqrt[e*x])/(Sqrt[e]*Sqrt[b*c - b*d*x])])/(Sqrt[d 
]*Sqrt[e])))/(Sqrt[c + d*x]*Sqrt[b*c - b*d*x])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 0]
 

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 140

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*d^(m + n)*f^p   Int[(a + b*x)^(m - 1)/(c + d*x)^m, x] 
, x] + Int[(a + b*x)^(m - 1)*((e + f*x)^p/(c + d*x)^m)*ExpandToSum[(a + b*x 
)*(c + d*x)^(-p - 1) - (b*d^(-p - 1)*f^p)/(e + f*x)^p, x], x] /; FreeQ[{a, 
b, c, d, e, f, m, n}, x] && EqQ[m + n + p + 1, 0] && ILtQ[p, 0] && (GtQ[m, 
0] || SumSimplerQ[m, -1] ||  !(GtQ[n, 0] || SumSimplerQ[n, -1]))
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 586

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_) 
, x_Symbol] :> Simp[(a + b*x^2)^FracPart[p]/((c + d*x)^FracPart[p]*(a/c + ( 
b*x)/d)^FracPart[p])   Int[(e*x)^m*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && EqQ[b*c^2 + a*d^2, 0]
 

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.21

method	result	size
default	\(\frac {\sqrt {b \left (-d^{2} x^{2}+c^{2}\right )}\, x b e \left (\arctan \left (\frac {\sqrt {b d e}\, \left (-2 d x +c \right )}{2 d \sqrt {\left (-d x +c \right ) b e x}}\right ) \sqrt {2}\, \sqrt {-\frac {b \,c^{2} e}{d}}\, d -2 \sqrt {b d e}\, \ln \left (\frac {3 b c d e x -b \,c^{2} e +2 \sqrt {2}\, \sqrt {-\frac {b \,c^{2} e}{d}}\, \sqrt {\left (-d x +c \right ) b e x}\, d}{d x +c}\right ) c \right ) \sqrt {2}}{2 \sqrt {e x}\, \sqrt {d x +c}\, \sqrt {\left (-d x +c \right ) b e x}\, \sqrt {b d e}\, d \sqrt {-\frac {b \,c^{2} e}{d}}}\)	\(181\)

Input:

int((-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(1/2)/(d*x+c)^(3/2),x,method=_RETURNVER 
BOSE)
 

Output:

1/2*(b*(-d^2*x^2+c^2))^(1/2)*x*b*e*(arctan(1/2*(b*d*e)^(1/2)*(-2*d*x+c)/d/ 
((-d*x+c)*b*e*x)^(1/2))*2^(1/2)*(-1/d*b*c^2*e)^(1/2)*d-2*(b*d*e)^(1/2)*ln( 
(3*b*c*d*e*x-b*c^2*e+2*2^(1/2)*(-1/d*b*c^2*e)^(1/2)*((-d*x+c)*b*e*x)^(1/2) 
*d)/(d*x+c))*c)*2^(1/2)/(e*x)^(1/2)/(d*x+c)^(1/2)/((-d*x+c)*b*e*x)^(1/2)/( 
b*d*e)^(1/2)/d/(-1/d*b*c^2*e)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.69 \[ \int \frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\left [\frac {1}{2} \, \sqrt {2} \sqrt {-\frac {b}{d e}} \log \left (-\frac {17 \, b d^{3} x^{3} + 3 \, b c d^{2} x^{2} - 13 \, b c^{2} d x + b c^{3} + 4 \, \sqrt {2} \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (3 \, d^{2} x - c d\right )} \sqrt {d x + c} \sqrt {e x} \sqrt {-\frac {b}{d e}}}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) + \frac {1}{2} \, \sqrt {-\frac {b}{d e}} \log \left (-\frac {8 \, b d^{3} x^{3} - 7 \, b c^{2} d x + b c^{3} - 4 \, \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (2 \, d^{2} x - c d\right )} \sqrt {d x + c} \sqrt {e x} \sqrt {-\frac {b}{d e}}}{d x + c}\right ), -\sqrt {2} \sqrt {\frac {b}{d e}} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} \sqrt {e x} d \sqrt {\frac {b}{d e}}}{3 \, b d^{2} x^{2} + 2 \, b c d x - b c^{2}}\right ) + \sqrt {\frac {b}{d e}} \arctan \left (\frac {2 \, \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} \sqrt {e x} d \sqrt {\frac {b}{d e}}}{2 \, b d^{2} x^{2} + b c d x - b c^{2}}\right )\right ] \] Input:

integrate((-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(1/2)/(d*x+c)^(3/2),x, algorithm= 
"fricas")
                                                                                    
                                                                                    
 

Output:

[1/2*sqrt(2)*sqrt(-b/(d*e))*log(-(17*b*d^3*x^3 + 3*b*c*d^2*x^2 - 13*b*c^2* 
d*x + b*c^3 + 4*sqrt(2)*sqrt(-b*d^2*x^2 + b*c^2)*(3*d^2*x - c*d)*sqrt(d*x 
+ c)*sqrt(e*x)*sqrt(-b/(d*e)))/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) 
+ 1/2*sqrt(-b/(d*e))*log(-(8*b*d^3*x^3 - 7*b*c^2*d*x + b*c^3 - 4*sqrt(-b*d 
^2*x^2 + b*c^2)*(2*d^2*x - c*d)*sqrt(d*x + c)*sqrt(e*x)*sqrt(-b/(d*e)))/(d 
*x + c)), -sqrt(2)*sqrt(b/(d*e))*arctan(2*sqrt(2)*sqrt(-b*d^2*x^2 + b*c^2) 
*sqrt(d*x + c)*sqrt(e*x)*d*sqrt(b/(d*e))/(3*b*d^2*x^2 + 2*b*c*d*x - b*c^2) 
) + sqrt(b/(d*e))*arctan(2*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)*sqrt(e*x 
)*d*sqrt(b/(d*e))/(2*b*d^2*x^2 + b*c*d*x - b*c^2))]
 

Sympy [F]

\[ \int \frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\int \frac {\sqrt {- b \left (- c + d x\right ) \left (c + d x\right )}}{\sqrt {e x} \left (c + d x\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((-b*d**2*x**2+b*c**2)**(1/2)/(e*x)**(1/2)/(d*x+c)**(3/2),x)
 

Output:

Integral(sqrt(-b*(-c + d*x)*(c + d*x))/(sqrt(e*x)*(c + d*x)**(3/2)), x)
 

Maxima [F]

\[ \int \frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\int { \frac {\sqrt {-b d^{2} x^{2} + b c^{2}}}{{\left (d x + c\right )}^{\frac {3}{2}} \sqrt {e x}} \,d x } \] Input:

integrate((-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(1/2)/(d*x+c)^(3/2),x, algorithm= 
"maxima")
 

Output:

integrate(sqrt(-b*d^2*x^2 + b*c^2)/((d*x + c)^(3/2)*sqrt(e*x)), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (110) = 220\).

Time = 0.25 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.85 \[ \int \frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {e x} (c+d x)^{3/2}} \, dx=-\frac {b^{3} {\left (\frac {\sqrt {2} \sqrt {-b d e} c d \log \left (\frac {{\left | -4 \, \sqrt {2} b^{2} {\left | c \right |} {\left | d \right |} {\left | e \right |} + 6 \, b^{2} c d e + 2 \, {\left (\sqrt {-b d e} \sqrt {-{\left (d x + c\right )} b + 2 \, b c} - \sqrt {b^{2} c d e + {\left ({\left (d x + c\right )} b - 2 \, b c\right )} b d e}\right )}^{2} \right |}}{{\left | 4 \, \sqrt {2} b^{2} {\left | c \right |} {\left | d \right |} {\left | e \right |} + 6 \, b^{2} c d e + 2 \, {\left (\sqrt {-b d e} \sqrt {-{\left (d x + c\right )} b + 2 \, b c} - \sqrt {b^{2} c d e + {\left ({\left (d x + c\right )} b - 2 \, b c\right )} b d e}\right )}^{2} \right |}}\right )}{b^{2} {\left | c \right |} {\left | d \right |} {\left | e \right |}} - \frac {\sqrt {-b d e} \log \left ({\left (\sqrt {-b d e} \sqrt {-{\left (d x + c\right )} b + 2 \, b c} - \sqrt {b^{2} c d e + {\left ({\left (d x + c\right )} b - 2 \, b c\right )} b d e}\right )}^{2}\right )}{b^{2} e}\right )}}{{\left | b \right |} {\left | d \right |}} \] Input:

integrate((-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(1/2)/(d*x+c)^(3/2),x, algorithm= 
"giac")
 

Output:

-b^3*(sqrt(2)*sqrt(-b*d*e)*c*d*log(abs(-4*sqrt(2)*b^2*abs(c)*abs(d)*abs(e) 
 + 6*b^2*c*d*e + 2*(sqrt(-b*d*e)*sqrt(-(d*x + c)*b + 2*b*c) - sqrt(b^2*c*d 
*e + ((d*x + c)*b - 2*b*c)*b*d*e))^2)/abs(4*sqrt(2)*b^2*abs(c)*abs(d)*abs( 
e) + 6*b^2*c*d*e + 2*(sqrt(-b*d*e)*sqrt(-(d*x + c)*b + 2*b*c) - sqrt(b^2*c 
*d*e + ((d*x + c)*b - 2*b*c)*b*d*e))^2))/(b^2*abs(c)*abs(d)*abs(e)) - sqrt 
(-b*d*e)*log((sqrt(-b*d*e)*sqrt(-(d*x + c)*b + 2*b*c) - sqrt(b^2*c*d*e + ( 
(d*x + c)*b - 2*b*c)*b*d*e))^2)/(b^2*e))/(abs(b)*abs(d))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\int \frac {\sqrt {b\,c^2-b\,d^2\,x^2}}{\sqrt {e\,x}\,{\left (c+d\,x\right )}^{3/2}} \,d x \] Input:

int((b*c^2 - b*d^2*x^2)^(1/2)/((e*x)^(1/2)*(c + d*x)^(3/2)),x)
 

Output:

int((b*c^2 - b*d^2*x^2)^(1/2)/((e*x)^(1/2)*(c + d*x)^(3/2)), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\frac {\sqrt {e}\, \sqrt {b}\, \left (\int \frac {\sqrt {d x +c}\, \sqrt {-d^{2} x^{2}+c^{2}}}{\sqrt {x}\, c^{2}+2 \sqrt {x}\, c d x +\sqrt {x}\, d^{2} x^{2}}d x \right )}{e} \] Input:

int((-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(1/2)/(d*x+c)^(3/2),x)
 

Output:

(sqrt(e)*sqrt(b)*int((sqrt(c + d*x)*sqrt(c**2 - d**2*x**2))/(sqrt(x)*c**2 
+ 2*sqrt(x)*c*d*x + sqrt(x)*d**2*x**2),x))/e